I am trying to define the Galois group of complex functions. For example for sinus we have the product representation:

$$\sin(x) = x \prod_{k=1}^\infty \left( 1-\frac{x^2}{k^2\pi^2} \right)$$ The roots of $\sin$ are $0,\pm k \pi$, $k\in \mathbb{N}$. It is known that $\pi$ is transcendent, hence the field $\mathbb{Q}(\pi)$ is isomorphic as fields to $\mathbb{Q}(x)$. If we define $\sigma(\pi)=-\pi$ and extend it to a $\mathbb{Q}(\pi)/\mathbb{Q}$ automorphism, then the Galois Group of $\sin$ over $\mathbb{Q}$ should be $C_2$ the cyclic group.

1) Any suggestions how to define a Galois group of a complex function as a group of permutations of the roots of these functions?

2) Do you know of any other function where the "Galois group" as defined below gives a non trivial group other than $C_2$?

For instance: What is the group defined below, for the Ramanujan theta function at $b=1$ ( https://en.wikipedia.org/wiki/Infinite_product#Product_representations_of_functions) : $$f(a,1) = \prod_{n=0}^\infty (1+a^{n+1})(1+a^n)(1-a^{n+1})$$

It seems that it has something to do with roots of unity.

Preliminary definition: Let $\Lambda := \{ z \in \mathbb{C} | f(z) = 0 \}$. Let $\mathbb{Q}(\Lambda)$ be the smallest subfield of $\mathbb{C}$ which contains $\Lambda$. Define the Galois group of $f$ to be $$ Gal(f/\mathbb{Q}) := \{ \sigma \in Aut(\mathbb{Q}(\Lambda)/\mathbb{Q})| z \in \Lambda \rightarrow \sigma(z) \in \Lambda \text{ and } \sigma^{-1}(z) \in \Lambda\}$$ Notice that for polynomials, the last property is automatically fullfilled, as automorphisms commute with the polynomial in the following sense: $ \sigma(p(z)) = p(\sigma(z))$. The above example for $\sin$ shows that $C_2$ is a subgroup of the Galois group. Is it possible to prove given the above definition that $C_2$ equals the Galois group?

Edit: By the comments below, we should only consider the automorphisms of the field extension which leave $\mathbb{Q}$ unchanged and permute the roots of the function. In the example above we have for any root $ z = k\pi, k \in \mathbb{Z}$ that: $$ 0 = \sigma(0) = \sigma(\sin(z)) = \sin(z) = \sin(-z) = \sin(\sigma(z))$$ Hence $\sigma$ maps roots to roots and is an automorphism of $\mathbb{Q(\pi)}$ which leaves $\mathbb{Q}$ unchanged. I don't think one needs the continuity here. Or did I miss something? I think in general ( for $z \in \mathbb{Q}(\pi)$ ) we do not have $\sigma(\sin(z)) = \sin(\sigma(z))$, so @Micah is right with his argument. Don't know if this still gives something interesting or not. Anyway, it is not clear how $\sigma(\sin(z))$ should be defined, as it might be the case that $\sin(z)$ is not an element of $\mathbb{Q}(\pi)$ although $z$ is. (For example $\sin(\pi/4) = 1/\sqrt{2}$ is not an element of $\mathbb{Q}(\pi)$, so in general it does not make sense to define $\sigma$ on such elements.)

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    $\begingroup$ I think you run into topological issues. $\Bbb{Q}(\pi)$ has lots of abstract field automorphisms over $\Bbb{Q}$ which do not simply permute the roots of $\sin$ (e.g., by extending $\sigma_r(\pi)=r\pi$ for any rational $r$), because they affect the convergence of the product. On the other hand, if you let $\Bbb{Q}(\pi)$ inherit its topology from $\Bbb{R}$, it has no continuous field automorphisms. I guess the hope would be that you could find some other topology under which $\sigma$ is continuous but all my other $\sigma_r$ are not; I'm not sure whether this is a reasonable thing to hope for... $\endgroup$ – Micah May 13 '18 at 16:44
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    $\begingroup$ Ok thank you for your comment. Why do you think the automorphisms should be continous? $\endgroup$ – orgesleka May 13 '18 at 16:48
  • $\begingroup$ In order to do anything like Galois theory you need to be considering automorphisms $\rho$ which commute with $\sin$: $\sin(\rho x)=\rho(\sin x)$. If you express $\sin$ via that infinite product, this basically means you want to be able to move $\rho$ past the infinite product, which presumably means you want some topology in which $\rho$ is continuous and the product is convergent. $\endgroup$ – Micah May 13 '18 at 16:58
  • $\begingroup$ @Micah: The example $\sigma(\pi)=-\pi$ is just that: an example. If you know of a continuous automorphism which commutes with $\sin$ that would also be fine. Or maybe you have an idea which topology to choose? $\endgroup$ – orgesleka May 13 '18 at 17:16
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    $\begingroup$ Related: math.stackexchange.com/questions/94994/… $\endgroup$ – Watson May 13 '18 at 21:19

The question is ill-posed. Please find a reference on the subject of infinite Galois theory so you know what that subject is really about.

Since $\pi$ is transcendental over $\mathbf Q$, the purely transcendental extension $\mathbf Q(\pi)$ is not part of Galois theory. You can of course ask about the automorphisms of this field that fix $\mathbf Q$, but this is not called a Galois group. It is called the group of $\mathbf Q$-automorphisms of $\mathbf Q(\pi)$.

What you are trying to do is doomed to be uninteresting. Suppose there is a $\mathbf Q$-automorphism $\sigma$ of $\mathbf Q(\pi)$ with $\sigma(\pi) = a\pi$ for some integer $a$, necessarily nonzero (since $\sigma$ is injective). Then by $\mathbf Q$-linearity we have $\sigma^{-1}(\pi) =(1/a)\pi$, and your dream is that the roots of $\sin z$ are permuted by the automorphisms, so you need $1/a$ to be an integer. Thus $a = \pm 1$. Either $\sigma$ fixes $\pi$ or sends it to its negative. There is no $\mathbf Q$-automorphism of $\mathbf Q(\pi)$ permuting the roots of $\sin z$ that sends $\pi$ to any root besides $\pm \pi$. So there is really nothing interesting to do.

The study of the group of all automorphisms of a transcendental field extension has important connections to algebraic geometry, but trying to force the integer multiples of $\pi$ to behave like something in Galois theory looks like a dead end. You would be better off learning the well-developed theory of infinite-degree Galois extensions, which involves topological concepts in a surprising way.

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    $\begingroup$ Thank you for your answer. With $\sin$ was just an example. Maybe you know a non-trivial example? Also note that your answer is like saying just because the Galois group of $x^2+1$ over $\mathbb{Q}$ is $C_2$, the whole subject is uninteresting. $\endgroup$ – orgesleka May 14 '18 at 5:05
  • $\begingroup$ (+1) for the useful answer. Asuming I am interested in the group defined above in the question. Do you know of any function with "nontrivial" group? $\endgroup$ – orgesleka May 14 '18 at 5:16
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    $\begingroup$ Not at all. There are lots of interesting examples of finite Galois extensions, both abelian and nonabelian, and it can be extended in a beautiful way to infinite algebraic extensions. My point was that trying to shoehorn the integer multiples of $\pi$ into something like Galois theory does not really lead anywhere. $\endgroup$ – KCd May 14 '18 at 5:18
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    $\begingroup$ Complex-analytic functions with infinitely many roots are not part of Galois theory. If you want to see a framework where there are lots of automorphisms permuting roots of a power series, study $p$-adic analysis. $\endgroup$ – KCd May 14 '18 at 5:28
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    $\begingroup$ Still my question, which you haven't answered, if you know of a complex function which, from your perspective, has a non-trivial group? $\endgroup$ – orgesleka May 14 '18 at 5:47

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